Problem 47 - Precalculus Student Solutions Manual 5th

Question

Solve each quadratic inequality. Write each solution set in interval notation. $$x^{2} \leq 9$$

Step-by-Step Solution

Answer

The solution set is \([-3, 3]\).

Step 1: Express the inequality in standard form

The inequality is already in a standard form: \( x^2 \leq 9 \)

Step 2: Move all terms to one side of the inequality

Rearrange the inequality such that one side is zero:\( x^2 - 9 \leq 0 \)

Step 3: Factor the quadratic expression

Factor the quadratic equation on one side of the inequality:\( x^2 - 9 = (x - 3)(x + 3) \)

Step 4: Identify the critical points

Set each factor equal to zero and solve for \(x\): \( x - 3 = 0 \implies x = 3 \) \( x + 3 = 0 \implies x = -3 \)These points, \(-3\) and \(3\), are the critical points that divide the number line into intervals.

Step 5: Test the intervals

Select a test point from each interval and plug it back into the inequality \( (x - 3)(x + 3) \leq 0 \).- Interval 1: \(x < -3\), pick \(x = -4\) \(( -4 - 3)( -4 + 3) = ( -7)( -1) = 7 \) (greater than 0)- Interval 2: \( -3 \leq x \leq 3\), pick \(x = 0\)\( (0 - 3)(0 + 3) = ( -3)(3) = -9 \) (less than or equal to 0)- Interval 3: \(x > 3\), pick \(x = 4\) \((4 - 3)(4 + 3) = (1)(7) = 7 \) (greater than 0)

Step 6: Determine the solution intervals

The inequality \( (x - 3)(x + 3) \leq 0 \) holds true for\( -3 \leq x \leq 3 \). Therefore, the solution interval in interval notation is:\([-3, 3]\)

Key Concepts

Interval Notation

Interval notation is a way of writing subsets of the real number line. It uses intervals to describe the set of solutions for an inequality. In our quadratic inequality, we determined that the values of \( x \) between -3 and 3 satisfy the inequality \( x^2 \leq 9 \). We include the end points of the interval because the inequality is \( \leq \), which means 'less than or equal to'. Therefore, the solution is written in interval notation as \([-3, 3]\).
Here are some key points about interval notation:

• Square brackets [ ] are used to denote that the endpoints are included (closed interval).
• Parentheses ( ) indicate that the endpoints are not included (open interval).
• The set of all real numbers can be represented as \( (-\infty, \infty) \).

Understanding interval notation helps us clearly express the range of values that are solutions to an inequality.

Factoring

Factoring is a crucial step in solving quadratic inequalities. It involves breaking down a complex expression into simpler expressions whose product is the original expression. In this problem, we started with \( x^2 - 9 \), which we factored as \( (x - 3)(x + 3) \). This process follows from the difference of squares formula: \( a^2 - b^2 = (a - b)(a + b) \).
Key benefits of factoring:

• Identifies the roots or critical points of the quadratic equation.
• Converts the inequality into a form that is easier to solve.

After factoring, we set each factor to zero to find the critical points. In doing so, we found that \( x = 3 \) and \( x = -3 \) are the points where the expression changes sign.

Critical Points

Critical points are values of \( x \) where the factored expression equals zero. These points divide the number line into different intervals. For the inequality \( x^2 - 9 \leq 0 \), the critical points were found by setting each factor equal to zero:\[ x - 3 = 0 \implies x = 3 \ x + 3 = 0 \implies x = -3 \]These values, -3 and 3, create three intervals on the number line:

• Interval 1: \( x < -3 \)
• Interval 2: \( -3 \leq x \leq 3 \)
• Interval 3: \( x > 3 \)

Critical points are essential because they tell us where the expression changes sign. Understanding this helps in determining where the inequality holds true.

Testing Intervals

Testing intervals involves checking a number from each interval to determine where the inequality holds. For our inequality, we examined three intervals derived from the critical points -3 and 3:

• For \( x < -3 \), we picked \( x = -4 \).
  Plugging in: \( (-4 - 3)(-4 + 3) = 7 \). Since 7 is greater than 0, this interval does not satisfy \( (x - 3)(x + 3) \leq 0 \).
• For \( -3 \leq x \leq 3 \), we chose \( x = 0 \).
  Plugging in: \( (0 - 3)(0 + 3) = -9 \). Since -9 is less than or equal to 0, this interval satisfies the inequality.
• For \( x > 3 \), we selected \( x = 4 \).
  Plugging in: \( (4 - 3)(4 + 3) = 7 \). Because 7 is greater than 0, this interval does not satisfy the inequality.

Testing intervals ensures that we understand where the quadratic expression is less than or equal to zero. We then translate this information into our solution in interval notation.